Answer:
The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]
Explanation:
Given that,
Number of turns = 22 turns
Number of turns another coil = 330 turns
Length of solenoid = 21.0 cm
Diameter = 2.30 cm
Current in inner solenoid = 0.140 A
Rate = 1800 A/s
Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid
We need to calculate the magnetic flux
Using formula of magnetic flux
[tex]\phi=BA[/tex]
[tex]\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2[/tex]
Put the value into the formula
[tex]\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2[/tex]
[tex]\phi=11.486\times10^{-8}\ Wb[/tex]
Hence, The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]
